Question

Find the sum of a 9-term geometric sequence when the first term is 4 and the last term is 1,024

Answer 1

$a_1=4$
$a_2=ra_1=4r$
$a_3=ra_2=r^2a_1=4r^2$
...
$a_9=ra_8=\cdots=r^8a_1=4r^8$

So you have

$1024=4r^8\implies 256=r^8\implies r=2$

So the sum is

$4+8+16+\cdots+512+1024$
$4(1+2+4+\cdots+128+256)$
$4(2^0+2^1+2^2+\cdots+2^7+2^8)$
$4\displaystyle\sum_{i=1}^92^{i-1}$
$4\times\dfrac{1-2^9}{1-2}$
$2044$